42O
Square Shafts.
according to its distance from D, the point of greatest stress.
Thus, if the ring K L be projected to D N, and N E joined, the
length KM will represent the virtual length of the ring if fs be
constant. Treating every perpendicular similarly, we obtain the
curve E p D, and the shaded figure is the virtual stress area, or
area of equal stress. Now cut out a copy of the shaded figure in
thin cardboard, and, hanging loosely from a pin in two -different
positions, as at w, mark plumb lines from the pin in each case
upon the paper. The crossing point will be o, the centre of
gravity, or centre of all the stresses, and the arm = E R. Next,
find the area of the figure. Divide s into 10 parts, and measure
everything in terms of these parts. Divide r> T into i o parts, and
draw horizontals from the middle of each part; then measure
their intercepts on the figure. Adding all these figures
(•13, -44, -82, &c.) and dividing by 10 we get the mean width
2*445, or "2445 s- The height DT measures 22*12 parts, or
2*212 s, so the area = height x mean width, and
Moment of resistance of section =/s x stress area x arm
=fs x height x mean width x arm
=/s X 2'2I2 S X -2445 S X -435 S =/S('235*S)*
St. Venant shewed, however, in 1856, that the previous
methods (Coulomb's, ring theory) were not strictly applicable to
any but circular sections, and gave the following:
Moment of square section =^(-208 s3) or '88 of {^('235 s8) ]•
because the greatest stress occurs at the middle of each side. To
illustrate St. Venant, Thomson and Tait have imagined the shaft
to be a box full of liquid, which, if rotated, woukj leave the latter
behind somewhat, and the apices would cause two stresses—
tangential and centripetal—to act on the particles, the former
only being of momental, value.
* Generally Tm=^ -, Zt =-, and I=Zfc^, where I is the polar moment
of inertia (see p. 429),